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Thermal radiation

Apply emissivity and the Stefan–Boltzmann law to net thermal-radiation exchange, using clear signs and limiting cases.

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  • University Physics Year 1
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Why this matters + quick links

This page gives the UY1 working model/result for Radiation. You reuse it when you turn thermal intuition into calculable models (energy balances, process reasoning, and kinetic/statistical links), with careful sign/units checks.

At a glance

  • Prerequisites: absolute temperature, surface area, power.
  • Outcomes: compute emitted and net radiative heat transfer.
  • Key result: Q dot ₙₑₜ = εσ A(Tₛ⁴-Tₛᵤᵣ⁴).
  • Common trap: using Celsius instead of kelvin in T⁴ expressions.

Setup

Surface at absolute temperature Tₛ exchanges radiation with surroundings at Tₛᵤᵣ.

  • Emissivity 0 ≤ ε ≤ 1.
  • Stefan-Boltzmann constant σ = 5.67 × 10⁻⁸ W m⁻²K⁻⁴.

Physical interpretation:

  • ε measures how “black” a surface is: ε = 1 is an ideal blackbody; small ε means weak emission/absorption.
  • The T⁴ scaling means radiation becomes important at high absolute temperature even when convection is present.

Bridge note:

  • Conduction/convection require matter; radiation works across vacuum.
  • UY1 upgrade is quantitative net-exchange modeling with emissivity and fourth-power temperature dependence.

Core derivation / explanation

Emission from a gray surface:

Q dot ₑₘᵢₜ = εσ A Tₛ⁴.

Absorption from large isothermal surroundings:

Q dot _abs = εσ A Tₛᵤᵣ⁴.

Net from surface to surroundings:

Q dot ₙₑₜ = Q dot ₑₘᵢₜ-Q dot _abs = εσ A(Tₛ⁴-Tₛᵤᵣ⁴).

For small temperature difference around mean Tₘ, linearized form:

Q dot ₙₑₜ ≈ hᵣA(Tₛ-Tₛᵤᵣ), hᵣ ≈ 4εσ Tₘ³.

This allows combined convection+radiation resistance modeling.

Pitfalls (radiation heat transfer)
  • Always use kelvin in any T⁴ expression. - Distinguish emitted power Q dot ₑₘᵢₜ = εσ ATₛ⁴ from net exchange Q dot ₙₑₜ = εσ A(Tₛ⁴-Tₛᵤᵣ⁴). - The simple absorption form used here assumes “large isothermal surroundings” (effectively a view factor of 1). More complex geometries need view factors. - The linearized hᵣ form is only accurate for sufficiently small temperature differences about Tₘ.

Units check:

[σ T⁴ A] = (W m⁻²K⁻⁴)K⁴m² = W.

Physical sign/limit checks:

  • If Tₛ > Tₛᵤᵣ, net heat loss is positive outward.
  • If Tₛ = Tₛᵤᵣ, net exchange is zero.
  • Radiation grows rapidly with temperature because of T⁴ dependence.

Net radiative cooling of a surface

Worked example 1

Emission minus absorption

Problem

A gray matte surface has ε = 0.90, area 0.50 m² and temperature 400 K. It is fully surrounded by a large isothermal enclosure at 300 K. Find its net radiative heat-loss rate using σ = 5.67 × 10⁻⁸ W m⁻²K⁻⁴.
Show full solution

  1. Subtract absorbed radiation

    Method

    Use εσ A(Tₛ⁴-Tₛᵤᵣ⁴).

    Reason

    The enclosure also emits radiation toward the surface, so emission alone overestimates the net loss.

    Working

    Answer

    Q dot ₙₑₜ = 0.90(5.67 × 10⁻⁸)(0.50)(400⁴-300⁴) ≈ 4.47 × 10² W.

  2. State the outward heat rate

    Working

    Answer

    Q dot ₙₑₜ ≈ 4.47 × 10² W, about 4.5 × 10² W to two significant figures. It is positive outward because 400⁴ > 300⁴; the units are W, not J.

Practice set

Check your understanding 1

A black surface has area 1.2 m² and temperature 350 K inside a large isothermal enclosure at 300 K. Find its net radiative heat-loss rate using σ = 5.67 × 10⁻⁸ W m⁻²K⁻⁴.

Show hint
A black surface has ε = 1; subtract the fourth powers.
Show answer
Q dot ₙₑₜ = (5.67 × 10⁻⁸)(1.2)(350⁴-300⁴) ≈ 470 W outward. Do not use (350-300)⁴: the two emitted powers must be subtracted.

Check your understanding 2

For a gray surface in the same large enclosure, emissivity halves while area and both temperatures stay fixed. What happens to net radiative heat transfer?

Show hint
Identify the dependence on ε in the net-exchange law.
Show answer
It halves because Q dot ₙₑₜ ∝ ε. Both emission and absorption scale with emissivity in this gray-surface model.

Check your understanding 3

Why can polishing a hot metal surface reduce its radiative heat loss to cooler surroundings?

Show hint
Consider which surface property multiplies the radiation law.
Show answer
Polished metal usually has lower emissivity. For the same area and temperatures, lower ε reduces εσ A(Tₛ⁴-Tₛᵤᵣ⁴) and hence the net radiative loss.

Summary + next steps

  • Radiation transfer depends strongly on absolute temperature and emissivity.
  • Net exchange is a temperature-difference-in-fourth-powers law.
  • At high temperature, radiation can dominate over convection.

Next steps:

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Syllabus and review details

No official syllabus alignment is listed for this lesson.

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